Question

What is the weight of oxygen required for the complete combustion of. 2.8kg of ethylene

Answer 1

The chemical equation for the said reaction is,

C2H4+12×2+4=283O296→2CO2+2H2O

the gram molar weight of ethylene is= 12*2+1*4 g
                                                              =28g
the gram molar weight of oxygen      = 3*32 g
                                                              =96 g 
∵ The weight of oxygen required for complete combustion of 28 g ethylene         =96 g 
according to question,

∴ Weight of oxygen required for combustion of 2.8 kg ethylene    =96×2.8×100028×1000kg
 
   =9.6kg
That is the answer.
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Answer 2

Answer: 9600 g or 9.6 kg

Explanation: $C_2H_4 (g)+3O_2 (g)\rightarrow 2CO_2(g)+2H_2O(g)$

As can be seen from the given balanced equation:

1 mole of ethylene reacts with = 3 moles of  oxygen gas

1 mole of ethylene weigh= 28 g

3 moles of oxygen weigh=$3\times 32g=96g$

Thus 28 g of ethylene require= 96 g of oxygen gas

$2.8kg=2.8\times 1000=2800 g$of  ethylene require=[tex]\farc{96}{28}\times 2800g=9600g of oxygen gas